ACdream 1113 The Arrow (概率dp求期望)-优快云博客

本文探讨了一个关于掷骰子的数学问题，即掷出特定数值的期望次数，并提供了两种不同的代码实现方案。

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E - The Arrow

Time Limit:1000MS Memory Limit:64000KB 64bit IO Format:%lld & %llu

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Description

The history shows that We need heroes in every dynasty. For example, Liangshan Heroes. People hope that these heroes can punish the bad guys and recover the justice. Nowadays, we also need heroes to provent us from being chopped, or being attacked by a bomb.

Kuangbin is a very very very very very.... (very * 1e9 ) good boy, after he knows The Arrow, he want to be The Arrow of China. But he is also a little afraid of being killed by the bad guys. So he decides to throw dices to make the decision.

The dice is a cube with 1 2 3 4 5 6 on it's sides. When he throws a dice, every number is of the same probablity to appear. He will write down a number N in the paper at first, and then throw the dice. When the sum of the number he throwed is less than N, he will keep throwing. But if the sum exceeds N, this throwing does not count.

For example, If the sum is 5,and N is 6, if we throw 2, 5 + 2 > 6, then the sum keeps to be 5.

If he can end the throwing in a certain time, he will make the decision to become The Arrow.

Now , kuangbin wonders that what's the expectation of the time of throwing dices.

Input

First line, The number of cases t <= 100

In the next t lines, there will be t numbers.

every number is not bigger than 100000

Output

Each test output a number rounded to the second digit.

Sample Input

1
1

Sample Output

6.00
题意：给定一个数n。现有一个骰子，六个面分别是123456，掷骰子，记录下掷出来的数的总和以及掷的次数，求和为n的时候，掷的次数的期望为多少。如果掷的和超过n就保持原来的n，比如n=10，第一次掷5第二次掷6，和是11>10，n保持5不变。有T组数据，每组数据给定一个n。
题解：掷到123456的期望是6，掷到7的期望是掷到123456的期望总和除以6+1。给出优化前和优化后的代码：
优化前：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int main()
{
    int t;
    double dp[101005];
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=6;i++)
            dp[i]=6;
        for(int i=7;i<=n;i++)
        {
            dp[i]++; //又掷了一次
            for(int j=1;j<=6;j++)
                dp[i]=dp[i]+dp[i-j]/6;
        }
        printf("%.2lf\n",dp[n]);
    }
    return 0;
}

优化后：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=1e5+5;
double dp[maxn];
void get()
{
    memset(dp,0,sizeof(dp));
    for(int i=1; i<=6; i++)
        dp[i]=6;
    for(int i=7; i<=maxn; i++)
        dp[i]=dp[i-1]-dp[i-7]/6+dp[i-1]/6;
}
int main()
{
    get();
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        printf("%.2lf\n",dp[n]);
    }
    return 0;
}